Timeline for On the sum of uniform independent random variables
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jun 18, 2016 at 21:42 | comment | added | user31317 | Do you think we can do a coupling between $n$ and $n+1$ sums? | |
| Jun 18, 2016 at 21:34 | comment | added | user31317 | I see what you are saying. | |
| Jun 18, 2016 at 21:22 | comment | added | user31317 | Yes but we know that $p_i(1/2)=1/2$ for all $i$ | |
| Jun 18, 2016 at 21:20 | comment | added | Christian Remling | @user31317: I don't know how you want to argue, but you're missing something because I can easily cook up function with positive derivative for which the RHS is $<p_n(x)$ (see my comment again). | |
| Jun 18, 2016 at 21:13 | comment | added | user31317 | We use mean value theorem for the first order, RHS becomes $p_n(x)+p_{n}'(\xi) \frac{2x-1}{n}$ for some $\xi \in (x-x/n,x+x/n)$. We know that $p_n'(\xi)>0$. | |
| Jun 18, 2016 at 21:04 | comment | added | Christian Remling | This can't work because it doesn't address the key issue mentioned in my final paragraph. On a more technical level, we'd be dealing with $n\int p_n'(\xi_t)(t-x)\, dt$, and I can't tell what the sign of this is. | |
| Jun 18, 2016 at 21:03 | comment | added | user31317 | Also, can you explain how did you use taylor for (1)? | |
| Jun 18, 2016 at 21:01 | comment | added | user31317 | Why don't you go with first order approximation? | |
| Jun 18, 2016 at 20:54 | history | undeleted | Christian Remling | ||
| Jun 18, 2016 at 20:52 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Jun 18, 2016 at 20:40 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Jun 18, 2016 at 20:13 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Jun 18, 2016 at 20:06 | history | deleted | Christian Remling | via Vote | |
| Jun 18, 2016 at 20:01 | history | answered | Christian Remling | CC BY-SA 3.0 |